Let T: P2 -> R2 by T[p(x)]=\begin{bmatrix}p(0)\\p(1)\end{bmatrix}
Show T is a linear transformation. I understand that I have to implant f(x+y)=f(x)+f(y), but have no clue how to do this.
I also do not understand what I am row reducing to get the kernel.
Hint: What happens to $$ \alpha p_1(x)+\beta p_2(x) $$ under $T$, where $\alpha$ and $\beta$ are real numbers and $p_1$ and $p_2$ are polynomials of degree less than or equal to $2$?
For the second part, take $p(x)=ax^2+bx+c$ and force it to be in the kernel, i.e. $$ p(0)=c=0 $$ and so $$ p(1)=0\implies a+b=0\implies b=-a $$ so $p(x)=a(x^2-x)$, and $\ker(T)=\text{span}\{ x^2-x\}$, and your kernel is one dimensional, as you would expect from rank nullity.